Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 2x}{x + 3} = \dfrac{5x + 18}{x + 3}$
Explanation: Multiply both sides by $x + 3$ $ \dfrac{x^2 + 2x}{x + 3} (x + 3) = \dfrac{5x + 18}{x + 3} (x + 3)$ $ x^2 + 2x = 5x + 18$ Subtract $5x + 18$ from both sides: $ x^2 + 2x - (5x + 18) = 5x + 18 - (5x + 18)$ $ x^2 + 2x - 5x - 18 = 0$ $ x^2 - 3x - 18 = 0$ Factor the expression: $ (x - 6)(x + 3) = 0$ Therefore $x = 6$ or $x = -3$ At $x = -3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -3$, it is an extraneous solution.